What would be the volume of 6.00 g of helium gas at STP?

2 Answers
Jun 12, 2018

"33.6 L"33.6 L.

Explanation:

To answer this question, we'll need to use the Ideal Gas Law:

pV = nRTpV=nRT,
where pp is pressure, VV is volume, nn is the number of moles RR is the gas constant, and TT is temperature in Kelvin.

The question already gives us the values for pp and TT, because helium is at STP. This means that temperature is "273.15 K"273.15 K and pressure is "1 atm"1 atm.

We also already know the gas constant. In our case, we'll use the value of "0.08206 L atm/K mol"0.08206 L atm/K mol since these units fit the units of our given values the best.

We can find the value for nn by dividing the mass of helium gas by its molar mass:

n = "number of moles" = "mass of sample"/"molar mass"n=number of moles=mass of samplemolar mass
= "6.00 g"/"4.00 g/mol" = "1.50 mol"=6.00 g4.00 g/mol=1.50 mol

Now, we can just plug all of these values in and solve for VV:

pV = nRTpV=nRT

V = (nRT)/p = ("1.50 mol" xx "0.08206 L atm/K mol" xx "273.15 K")/"1 atm"V=nRTp=1.50 mol×0.08206 L atm/K mol×273.15 K1 atm

"= 33.6 L"= 33.6 L

Jun 12, 2018

33.6 litres

Explanation:

Assuming that helium behaves as an ideal gas, we know that a mole of ideal gas occupies 22.4 litres at STP.

The atomic weight of helium is 4 g/mol, so 6 g is 1.5 moles.

The volume occupied at STP, therefore, is 22.4 x 1.5 = 33.6 litres.